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2 years ago

Write the equation of a circle with a center at (12, 6) and a radius of 6.

Mathematics

1 answer:

uranmaximum [27]2 years ago

3 0

Answer:

(x-12)² + (y-6)² = 36 (Option C)

Step-by-step explanation:

use circle formula

(x-h)² + (y-k)²= r²

h= 12 and k= 6 and r= 6

(x-12)² + (y-6)² = 6²

6 squared = 36 (6·6)

(x-12)² + (y-6)² = 36

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Kazeer [188]

Answer:

(x-y)(x-y-3)

Step-by-step explanation:

(x-y)² - 3(x-y)

take (x-y) common,

(x-y)(x-y-3)

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3 years ago

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Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

$\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

but the integrand is even, so this is really just

$\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

Substitute x = 1/2 arccot(u/2), which transforms the integral to

$\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du$

There are lots of ways to compute this. What I did was to consider the complex contour integral

$\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz$

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

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which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

$\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}$

and it follows that

$\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}$

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1 year ago

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Olin [163]

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3 years ago

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2 years ago

Plz give 8th and 9th question

Neko [114]

Answer:

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Step-by-step explanation:

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2 years ago